What is the first d/(dt)(t^3-t) ?

The first d/(dt)(t^3-t) is 3t^2-1

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d/(dt)(t^3-t)

\frac{d}{d t} (t^{3} - t)

3 t^{2} - 1

Solution steps

\frac{d}{d t} (t^{3} - t)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d t} (t^{3}) - \frac{d t}{d t}

\frac{d}{d t} (t^{3}) = 3 t^{2}

\frac{d t}{d t} = 1

= 3 t^{2} - 1

\frac{d y}{d x} = x x - 15

\frac{6}{s} + 8 a \frac{1}{s ^{2} + 16}

\frac{d}{d x} (- 3 x y)

\frac{d}{d x} y + y = 7

\int \frac{1}{x 9 + 4 x ^{2}} d x